# Toronto Math Forum

## MAT244-2013F => MAT244 Math--Tests => MidTerm => Topic started by: Victor Ivrii on October 09, 2013, 07:20:21 PM

Title: MT, P1
Post by: Victor Ivrii on October 09, 2013, 07:20:21 PM
Solve the following initial value problem:
\begin{equation*}
\end{equation*}
Title: Re: MT, P1
Post by: Yangming Cai on October 09, 2013, 10:00:17 PM
Title: Re: MT, P1
Post by: Xiaozeng Yu on October 09, 2013, 10:18:28 PM
1
Title: Re: MT, P1
Post by: Xuewen Yang on October 09, 2013, 10:47:45 PM
For Xiaozeng Yu,

Is there a problem with the sign change? from "plus" (2nd row) to "minus" (3rd row) ?
Title: Re: MT, P1
Post by: Xiaozeng Yu on October 09, 2013, 10:52:39 PM
For Xiaozeng Yu,

Is there a problem with the sign change? from "plus" (2nd row) to "minus" (3rd row) ?

no mistake. $d(\ln(1-y))/dt = -1/(1-y)$.
Title: Re: MT, P1
Post by: Xuewen Yang on October 09, 2013, 10:57:34 PM
For Xiaozeng Yu,

Is there a problem with the sign change? from "plus" (2nd row) to "minus" (3rd row) ?

no mistake. d(ln(1-y))/dt = -1/1-y

You are right! My bad  :P
Title: Re: MT, P1
Post by: Victor Ivrii on October 10, 2013, 06:03:59 AM
And where this solution is defined? $\DeclareMathOperator{\arctanh}{arctanh}$

BTW, when collecting MT I have seen this problem and in many papers integration was atrocious.

As long as $|y|<1$ we have $\int \frac{dy}{1-y^2}=\arctanh (y)$ (inverse hyperbolic function $\tanh$) and then
$y=\tanh (t)=\frac{{e^t}-e^{-t}}{e^{t}+e^{-t}}$ (similar to $\int \frac{dy}{1+y^2}=\arctan(y)$).

Title: Re: MT, P1
Post by: Xiaozeng Yu on October 10, 2013, 09:53:07 AM
And where this solution is defined? $\DeclareMathOperator{\arctanh}{arctanh}$

BTW, when collecting MT I have seen this problem and in many papers integration was atrocious.

As long as $|y|<1$ we have $\int \frac{dy}{1-y^2}=\arctanh (y)$ (inverse hyperbolic function $\tanh$) and then
$y=\tanh (t)=\frac{{e^t}-e^{-t}}{e^{t}+e^{-t}}$ (similar to $\int \frac{dy}{1+y^2}=\arctan(y)$).

hmm...yea...-1<y<1  T.T i feel i'm screwed...
Title: Re: MT, P1
Post by: Victor Ivrii on October 10, 2013, 10:51:37 AM
And where this solution is defined?
hmm...yea...-1<y<1

This is a range of $y$ (what values it takes), I asked about domain--for which $t$ it is defined.

So, my question about domain is pending
Title: Re: MT, P1
Post by: Xiaozeng Yu on October 10, 2013, 10:51:42 PM
t is defined everywhere...
Title: Re: MT, P1
Post by: Victor Ivrii on October 11, 2013, 12:19:32 AM
$t$ is defined everywhere...

You think right but write incorrectly: Solution $y(t)$ is defined everywhere.

Title: Re: MT, P1
Post by: Xiaozeng Yu on October 11, 2013, 10:05:54 AM
thank u prof.